This tutorial shows you:

Note on copying & pasting code from the PDF version of this tutorial: Please note that you may run into trouble if you copy & paste code from the PDF version of this tutorial into your R script. When the PDF is created, some characters (for instance, quotation marks or indentations) are converted into non-text characters that R won’t recognize. To use code from this tutorial, please type it yourself into your R script or you may copy & paste code from the source file for this tutorial which is posted on my website.

Note on R functions discussed in this tutorial: I don’t discuss many functions in detail here and therefore I encourage you to look up the help files for these functions or search the web for them before you use them. This will help you understand the functions better. Each of these functions is well-documented either in its help file (which you can access in R by typing ?ifelse, for instance) or on the web. The Companion to Applied Regression (see our syllabus) also provides many detailed explanations.

As always, please note that this tutorial only accompanies the other materials for Day 10 and that you are expected to have worked through the reading for that day before tackling this tutorial.

This week’s tutorial focuses on violations of the OLS assumptions. As a reminder, the assumptions you’ve encountered are:

For violations of most of these assumptions, investigating the residuals of a regression model is a good start. Hence this tutorial will focus on residuals quite a bit. Residuals (or errors) are the differences between the observed values of \(y\) and the predicted, or fitted, values for \(y\), also known as \(\hat{y}\):

\[ \varepsilon_i = y_i - \hat{y}_i \]

You recall that we assume these residuals to be distributed normally with a mean of 0 and a constant variance. That is, the spread of the residuals around the predicted values is the same across the range of predicted values.

Non-normally distributed errors

You read in section 12.1 of AR that non-normally distributed errors do not compromise the validity of OLS estimates in large samples, but they may give you a reason to improve your model. There are two simple graphical ways to diagnose whether the residuals are not normally distributed, and you’re already familiar with these methods from the tutorial for Day 3.

For an example, we’ll use data from the Canadian Survey of Labour and Income Dynamics, for the province of Ontario. Observations are individuals from Ontario who responded to the 1994 wave of the survey. These data are provided as part of the “car” package, but I uploaded them to my website as well. The following variables are included:

slid.dat <- rio::import("http://www.jkarreth.net/files/car_SLID.csv")
summary(slid.dat)
##      wages          education          age            sex           
##  Min.   : 2.300   Min.   : 0.00   Min.   :16.00   Length:7425       
##  1st Qu.: 9.235   1st Qu.:10.30   1st Qu.:30.00   Class :character  
##  Median :14.090   Median :12.10   Median :41.00   Mode  :character  
##  Mean   :15.553   Mean   :12.50   Mean   :43.98                     
##  3rd Qu.:19.800   3rd Qu.:14.53   3rd Qu.:57.00                     
##  Max.   :49.920   Max.   :20.00   Max.   :95.00                     
##  NA's   :3278     NA's   :249                                       
##    language        
##  Length:7425       
##  Class :character  
##  Mode  :character  
##                    
##                    
##                    
##

This dataset contains 5 variables:

Variable Description
wages Composite hourly wage rate from all jobs in Canadian dollars.
education Number of years of schooling.
age Age in years
sex A categorical variable: Female, Male.
language A categorical variable: English, French, Other.

As often before, we’d like to predict hourly wages (wages) with education and, in this case, other variables. To do so, we can fit an OLS regression model as usual:

mod <- lm(wages ~ education + age + sex, data = slid.dat)
summary(mod)
## 
## Call:
## lm(formula = wages ~ education + age + sex, data = slid.dat)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -26.111  -4.328  -0.792   3.243  35.892 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -7.905243   0.607771  -13.01   <2e-16 ***
## education    0.918735   0.034514   26.62   <2e-16 ***
## age          0.255101   0.008634   29.55   <2e-16 ***
## sexMale      3.465251   0.208494   16.62   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.602 on 4010 degrees of freedom
##   (3411 observations deleted due to missingness)
## Multiple R-squared:  0.2972, Adjusted R-squared:  0.2967 
## F-statistic: 565.3 on 3 and 4010 DF,  p-value: < 2.2e-16

To investigate the residuals of this regression, we can use the resid() or (equivalently) residuals() functions.

summary(resid(mod))
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -26.1100  -4.3280  -0.7925   0.0000   3.2430  35.8900

But these quantities don’t tell us much about the distribution of the residuals.

Density plot

Hence, a density plot is often useful. You’re already familiar with density plots (and histograms) from Day 3.

plot(density(resid(mod)))

Last week (Day 9), you learned that normalizing/standardizing residuals often is useful because studentized residuals because they follow a t-distribution. You can obtain the studentized residuals using the rstudent() function.

summary(rstudent(mod))
##      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
## -3.968000 -0.656000 -0.120100  0.000099  0.491500  5.460000
plot(density(rstudent(mod)))

With your knowledge of the normal (and t-) distributions, you can easily tell when the (studentized) residuals are not normally distributed. In this case, you’ll note a slight “dent” in the right half of this density plot.

Quantile-comparison plots for residuals

A second graphical test for normality you’re already familiar with plots the actual percentiles of the residuals against theoretical percentiles of a normal distribution. The familiar qqnorm() function does this for you, in combination with the qqline() function.

qqnorm(rstudent(mod))
qqline(rstudent(mod))

See section 12.1 in AR and the tutorial from Day 3 for more detailed guidance on how to interpret these figures. In the quantile-comparison plot above, you probably notice that the tails of the distribution of the residuals deviates from the normal line, especially on the upper end. One reason for this is that our model is not predicting observations on the upper end of \(y\) (wages) well. As you’ll recall, wages or incomes are often heavily skewed. Let’s verify this:

hist(slid.dat$wages)

Transforming variables

In this case, it would be useful to create a transformed version of the outcome variable, wages, using the natural log. Recall that the log of 0 is undefined, though, so you must check if wages contains any values of 0 first. If it does, you can add a very small positive value (depending on the scale of the variable).1 In this case, wages does not contain 0s, so taking the log without any changes is not problematic.

summary(slid.dat$wages)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
##   2.300   9.235  14.090  15.550  19.800  49.920    3278
slid.dat$wages.ln <- log(slid.dat$wages)

Fitting the model with the logged outcome variable instead also returns a less “concerning” quantile-comparison plot, at least at the upper tail:

mod.ln <- lm(wages.ln ~ education + age + sex, data = slid.dat)
summary(mod.ln)
## 
## Call:
## lm(formula = wages.ln ~ education + age + sex, data = slid.dat)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.36252 -0.27716  0.01428  0.28625  1.56588 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 1.1168632  0.0385480   28.97   <2e-16 ***
## education   0.0552139  0.0021891   25.22   <2e-16 ***
## age         0.0176334  0.0005476   32.20   <2e-16 ***
## sexMale     0.2244032  0.0132238   16.97   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4187 on 4010 degrees of freedom
##   (3411 observations deleted due to missingness)
## Multiple R-squared:  0.3094, Adjusted R-squared:  0.3089 
## F-statistic: 598.9 on 3 and 4010 DF,  p-value: < 2.2e-16
qqnorm(rstudent(mod.ln))
qqline(rstudent(mod.ln))

A density plot of the studentized residuals reveals the same, as the “dent” in the right has now disappeared:

plot(density(rstudent(mod.ln)))

However, note that your interpretation of these estimates has changed with the transformation of the outcome variable. Each \(\beta\) now expresses the “effect” of a one-unit increase in \(x\) on the log transformation of \(y\). This requires some extra work and care on interpreting and presenting your results.

Nonconstant error variance (heteroskedasticity)

A potentially bigger concern about inferences from OLS models comes from violations of the assumption of constant error variance. Under heteroskedasticity, the standard errors from OLS are inaccurate. You’ll find more details about this problem on p. 272 and pp. 276-7 in AR. Broadly speaking, you should consider heteroskedasticity as an indicator for a mis-specified model: your model may be missing an important part of the explanation for variation in your outcome variable \(y\).

Residual plots

You’re already familiar with residual plots from bivariate regression: you plotted residuals (on the y-axis) against the predictor (on the x-axis). With multiple regression, you can do the same for each predictor.

Plot residuals against each predictor

When you fit a model in R, the model object (the result of your call to lm()) conveniently stores the variables used in the model in the sub-object model. This object is a dataframe. This is convenient for you because the number of observations in a model might not be the same as the number of observations in your data. R (and any other statistical software) automatically removes observations from a regression model if any observation has missing values on any of the outcome or explanatory variables.

To access the data used for estimating the model, access the “model matrix” by calling modelname$model, where modelname is the object you created when fitting the regression model. In our case, this is mod.ln:

summary(mod.ln$model)
##     wages.ln        education          age            sex           
##  Min.   :0.8329   Min.   : 0.00   Min.   :16.00   Length:4014       
##  1st Qu.:2.2216   1st Qu.:12.00   1st Qu.:28.00   Class :character  
##  Median :2.6476   Median :13.00   Median :36.00   Mode  :character  
##  Mean   :2.6193   Mean   :13.34   Mean   :37.09                     
##  3rd Qu.:2.9829   3rd Qu.:15.10   3rd Qu.:46.00                     
##  Max.   :3.9104   Max.   :20.00   Max.   :69.00

You can now use these variables to plot each predictor against the residuals. A scatterplot of residuals and the binary predictor gender is not all that useful, so I’m only plotting the two continuous predictors:

par(mfrow = c(1, 2))
plot(x = mod.ln$model$education, y = resid(mod.ln))
plot(x = mod.ln$model$age, y = resid(mod.ln))

par(mfrow = c(1, 1))

Plot residuals against fitted values

You can also plot the residuals against the fitted values of \(y\). My writing here supplements what you read in section 12.2.1 in AR, using similar data and the same empirical model.

The latter is easily accomplished using the resid() (or rstudent()) and fitted.values() functions. I’m returning to our model with the log-transformed version of the wages.

plot(x = fitted.values(mod.ln), y = rstudent(mod.ln))

We see here that the variance of the residuals is smaller at lower fitted values and then increases somewhat. This may indicate that the residuals are themselves a function of the fitted values. You should use this as a prompt to investigate this model further.

In addition to plotting residuals against fitted values, you can use the convenient residualPlots() function from the “car” package to plot residuals against each individual predictor. The default function usually gives you a useful diagnostic plot, but see the help file at ?residualPlots for customization options. This function also returns test statistics for the curvatore of the fitted line in each residual plots involving continuous predictors. Notice that the plot on the right reproduces the plot of residuals vs. fitted values that you just made by hand, and the two plots on the left are identical to what you just plotted yourself in the previous section.

library(car)
residualPlots(mod.ln, layout = c(1, 3), type = "rstudent")

##            Test stat Pr(>|t|)
## education      3.751        0
## age          -21.621        0
## Tukey test    -3.931        0

Nonlinearity

A very fundamental assumption of linear regression (our assumption “A1”) is that the data-generating process is linear: each predictor \(x\) is linearly related to \(y\). This means that a one-unit increase in \(x\) is associated with the same increase in \(y\) regardless of the value of \(x\). Revisiting our original model:

summary(mod)
## 
## Call:
## lm(formula = wages ~ education + age + sex, data = slid.dat)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -26.111  -4.328  -0.792   3.243  35.892 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -7.905243   0.607771  -13.01   <2e-16 ***
## education    0.918735   0.034514   26.62   <2e-16 ***
## age          0.255101   0.008634   29.55   <2e-16 ***
## sexMale      3.465251   0.208494   16.62   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.602 on 4010 degrees of freedom
##   (3411 observations deleted due to missingness)
## Multiple R-squared:  0.2972, Adjusted R-squared:  0.2967 
## F-statistic: 565.3 on 3 and 4010 DF,  p-value: < 2.2e-16

We would conclude that one additional year of education is associated with about 1 Canadian dollar (\(0.91\)) additional hourly wage. Moving from 5 years of education to 6 years yields the same increase in wage as moving from 19 to 20 years of education.

It should be obvious that this assumption is often not realistic. For instance, you could think of the idea of diminishing returns.2

Component-plus-residual plots

To check this assumption, you can create component-plus-residual plots as described in section 12.3.1 of AR. The “car” package offers the crPlots() function for this purpose. Let’s return to our model with the log-transformed wages outcome variable and create component-plus residual plots for the continuous predictors.

crPlots(mod.ln, layout = c(1, 3))

I can also use the terms = argument to remove a component-residual plot for the binary sex variable.

crPlots(model = mod.ln, terms = ~ . - sex, layout = c(1, 2))

Please refer to section 12.3.1 of AR for a thorough discussion of this example.

Collinearity

We’ve already encountered the concept of collinearity (AKA multicollinearity). You recall from the formula for the standard error of \(\hat{\beta}\) in multiple regression that strong associations between predictors will increase standard errors, and therefore increase the probability of a type-II error (failing to reject a false null hypothesis).

Variance-inflation factors

You could of course simply examine the correlation between predictors, but that ignores the fact that multicollinearity depends upon the association between possibly more than two predictors, as you saw in our calculation of \(R^2_j\), the \(R^2\)-value from the regression of one \(x_j\) on all other predictors. Therefore, a common diagnostic is the variance-inflation factor (VIF). The VIF for predictor \(x_j\) is a transformation of \(R^2_j\):

\[ \text{VIF}_j = \frac{1}{1 - R^2_j} \]

In R, the “car” package offers the function vif() to calculate the VIF for each predictor. Returning to our example:

vif(mod.ln)
## education       age       sex 
##  1.012179  1.011613  1.000834

AR wisely does not provide “critical values” for VIFs, and you should read AR’s discussion of multicollinearity carefully. Multicollinearity is often less of a problem than the suggested “fixes”. Let’s briefly investigate how the VIFs look for a model with data that we simulate to have highly correlated predictors. To this end, we make \(x_2\) a function of \(x_1\) and some random noise: \(x_2 = x1 + \mathcal{N}(0, 1)\)

set.seed(123)
n.sample <- 200
x1 <- rnorm(n.sample, 2, 5)
x2 <- x1 + rnorm(n.sample, 0, 1)
x3 <- runif(n.sample, min = -10, max = 10)
a <- 0.2
b1 <- 1.1
b2 <- 0.2
b3 <- -0.7
e <- rnorm(n.sample, 0, 10)
y <- a + b1 * x1 + b2 * x2 + b3 * x3 + e
mod.sim <- lm(y ~ x1 + x2 + x3)
summary(mod.sim)
## 
## Call:
## lm(formula = y ~ x1 + x2 + x3)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -27.0679  -6.7968  -0.0854   6.8281  25.6651 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.836708   0.759384  -1.102   0.2719    
## x1           1.343254   0.717766   1.871   0.0628 .  
## x2          -0.001452   0.705987  -0.002   0.9984    
## x3          -0.737298   0.121728  -6.057 6.96e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 9.903 on 196 degrees of freedom
## Multiple R-squared:  0.3842, Adjusted R-squared:  0.3748 
## F-statistic: 40.76 on 3 and 196 DF,  p-value: < 2.2e-16
vif(mod.sim)
##        x1        x2        x3 
## 23.248488 23.231691  1.003858
sqrt(vif(mod.sim))
##       x1       x2       x3 
## 4.821669 4.819926 1.001927
mod.sim2 <- lm(y ~ x1 + x3)
summary(mod.sim2)
## 
## Call:
## lm(formula = y ~ x1 + x3)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -27.0696  -6.7984  -0.0867   6.8293  25.6630 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -0.8368     0.7565  -1.106     0.27    
## x1            1.3418     0.1486   9.030  < 2e-16 ***
## x3           -0.7373     0.1213  -6.080 6.14e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 9.878 on 197 degrees of freedom
## Multiple R-squared:  0.3842, Adjusted R-squared:  0.378 
## F-statistic: 61.46 on 2 and 197 DF,  p-value: < 2.2e-16
vif(mod.sim2)
##       x1       x3 
## 1.001483 1.001483

Robust standard errors

You’re learning today (again) that non-constant error variance affects foremost the standard errors of a regression model. This would suggest that if standard errors can be corrected, your OLS estimates are still useful (since they are unbiased anyway, and with corrected standard errors, your inferences and significance tests of \(\hat{\beta}\) are also “corrected”.) Therefore it might not come as a surprise that econometricians have suggested corrections for the standard error in the presence of heteroskedasticity. One of these corrections goes by the name of the “heteroskedasticity-consistent standard error” or “Huber-White standard error” or “robust standard error.” Sometimes different authors refer to different implementations of this idea, but the logic is usually the same. It is explained in section 12.2.3 of AR (which was not required for today).

Before we advance, I’d like to emphasize that robust standard errors are often not an optimal “fix.” Heteroskedasticity often suggests deeper problems with your model specification; perhaps the model is missing an important predictor, or the outcome or one of the predictor variables should be transformed. Today’s applied reading, King and Roberts (2015) “How Robust Standard Errors Expose Methodological Problems They Do Not Fix, and What to Do About it” explains this in an accessible way.

With the assumption of homoskedasticity (constant error variance), we stipulate that the errors of each observation have, on average, the same variance. That is, no one observation should have a bigger expected error than any other observation. When this assumption is violated (i.e., when you diagnose heteroskedasticity), errors do not exhibit the same average variance anymore. Per White (1980), robust standard errors take this non-constant variance into account and adjust coefficient standard errors for it.

The idea behind robust standard errors is to add to the sampling variance of the coefficient, \(V(\hat{\beta})\) a correction that takes into account the non-constant variance of the residuals \(\varepsilon_i\). In section 12.2.3 of AR you see the equation behind this. In the calculation of \(V(\hat{\beta})\), robust standard errors use \(\text{diag}\{\frac{E_1^2}{(1 - h_1)^2}, \ldots \frac{E_n^2}{(1 - h_n)^2}\}\) instead of \(\text{diag}\{E_1^2, \ldots E_n^2\}\). In other words, the standard errors of the coefficients are adjusted for the hat-values (recall hat-valeus from Day 8).

To visualize this, consider equation 12.4 in AR. Recall that the variance-covariance matrix contains the standard errors of coefficients in its diagonal (sqrt(diag(vcov(mod))))). Under homoskedasticity, the error variance is \(\sigma_\varepsilon^2\) for all observations:

\[ \widehat{\mathbf{\Sigma}} = \begin{bmatrix} \sigma_\varepsilon^2 & 0 & 0 & \cdots & 0 \\ 0 & \sigma_\varepsilon^2 & 0 & \cdots & 0 \\ 0 & 0 & \sigma_\varepsilon^2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & 0 \\ 0 & 0 & 0 & \cdots & \sigma_\varepsilon^2 \\ \end{bmatrix} \]

and the variance-covariance matrix can be expressed as:

\[ \text{VarCov}(\beta) = \sigma_{\varepsilon}^2 (\mathbf{X}'\mathbf{X})^{-1} \]

Under heteroskedasticity, \(\sigma_\varepsilon^2\) for all observations does not hold anymore: different observations \(i\) have different error variances \(\sigma_{\varepsilon}^2\). These error variances for each observation \(\sigma_1^2\), \(\sigma_2^2\), etc. can be thought of as the diagonal of an “error matrix”:3

\[ \widehat{\mathbf{\Sigma}} = \begin{bmatrix} \sigma_1^2 & 0 & 0 & \cdots & 0 \\ 0 & \sigma_2^2 & 0 & \cdots & 0 \\ 0 & 0 & \sigma_3^2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & 0 \\ 0 & 0 & 0 & \cdots & \sigma_n^2 \\ \end{bmatrix} \]

In this case, the variance-covariance matrix uses the above “error matrix” to adjust standard errors for this non-constant error variance:

\[ \text{VarCov}_{\text{robust}}(\beta) = (\mathbf{X}'\mathbf{X})^{-1} \mathbf{X}'\widehat{\mathbf{\Sigma}}\mathbf{X} (\mathbf{X}'\mathbf{X})^{-1} \]

I’m providing you with a small R function, robSE(), to calculate robust standard errors for OLS estimates. I’m providing this function on GitHub. You can either copy and paste it into R from there, and then use it with an lm object. You have to paste the function once in your R session before you use it. Alternatively, you can read it in using the source_url() function from the “devtools” package. I show in this demo https://github.com/jkarreth/JKmisc/blob/master/robclSE.demo.R how that’s done.

For this tutorial, I paste in the robSE() function after copying it from my GitHub page.

robSE <- function(mod){

  require(lmtest, quiet = TRUE)

    X.mat <- model.matrix(mod) # note that X includes 1 for the intercept
    y <- mod$mod[1] # Response variable
    e <- resid(mod) # Vector of residuals
    n <- dim(X.mat)[1] # Number of observations
    k <- dim(X.mat)[2] # Number of predictors (including intercept)
    
    vcov.mat <- vcov(mod)   # Variance-covariance matrix
    SE <- sqrt(diag(vcov.mat)) # Standard errors
    E.mat <- matrix(0, ncol = n, nrow = n)  # n-by-n Matrix of residuals
    diag(E.mat) <- e^2  # Squared residuals in the diagonal, 0 in the off-diagonal
    
    sum.mat <- t(X.mat) %*% E.mat %*% X.mat # the middle part of eq. 12.5
    
    vcov.mat.rob <- abs(solve(t(X.mat) %*% X.mat) %*% sum.mat %*% solve(t(X.mat) %*% X.mat)) # eq. 12.5
    
    rob.SE <- sqrt(diag(vcov.mat.rob))
    ## Add the degrees-of-freedom correction
    dfc <- sqrt(nrow(X.mat)) / sqrt(nrow(X.mat) - ncol(X.mat))
    rob.SE.dfc <- dfc * sqrt(diag(vcov.mat.rob))
    return(rob.SE.dfc)

    }

A couple of notes:

Try it ourself with our initial model (before transforming wages). Here, I’m extracting only the standard errors as the second column ([, 2]) of the matrix of coefficients, standard errors, t-values, and p-values of the model summary [summary(mod)$coefficients]:

summary(mod)$coefficients[, 2]
## (Intercept)   education         age     sexMale 
## 0.607770973 0.034514202 0.008634409 0.208494153
robSE(mod)
## (Intercept)   education         age     sexMale 
## 0.651790303 0.038671314 0.009057728 0.208782964

To put both “original” and “robust” standard errors side-by-side, you can use the screenreg() function, as explained in more detail at https://github.com/jkarreth/JKmisc/blob/master/robclSE.demo.R.

mod_robSE <- robSE(mod)
mod_robSE_pval <- ((1 - pnorm(abs(coef(mod) / robSE(mod)))) * 2)
library(texreg)
screenreg(list(mod, mod), 
  override.se = list(summary(mod)$coefficients[, 2], mod_robSE),
  override.pvalues = list(summary(mod)$coefficients[, 4], mod_robSE_pval),
  custom.model.names = c("Original SEs", "Robust SEs"),
  digits = 5
  )
## 
## ===========================================
##              Original SEs    Robust SEs    
## -------------------------------------------
## (Intercept)    -7.90524 ***    -7.90524 ***
##                (0.60777)       (0.65179)   
## education       0.91873 ***     0.91873 ***
##                (0.03451)       (0.03867)   
## age             0.25510 ***     0.25510 ***
##                (0.00863)       (0.00906)   
## sexMale         3.46525 ***     3.46525 ***
##                (0.20849)       (0.20878)   
## -------------------------------------------
## R^2             0.29720         0.29720    
## Adj. R^2        0.29668         0.29668    
## Num. obs.    4014            4014          
## RMSE            6.60189         6.60189    
## ===========================================
## *** p < 0.001, ** p < 0.01, * p < 0.05

You can see that the robust standard errors are pretty close to the original standard errors in this case.

Note: I’m introducing robust standard errors here without endorsement. But, depending on your subfield, you will encounter them in published work, and I want to make sure you see the mechanics of them and how to obtain them in R.

  1. A good discussion of this practice can be found at http://stats.stackexchange.com/questions/1444/how-should-i-transform-non-negative-data-including-zeros.↩

  2. See, for instance, http://www.investopedia.com/terms/l/lawofdiminishingmarginalreturn.asp.↩

  3. Some of this content borrows from Matt Golder’s notes.↩